Topic 1.4: Mass Percent and Mixture Composition Quiz

Boost your AP Chemistry score with this Topic 1.4 practice quiz. Test your understanding of mixture composition, mass percent, and elemental analysis for exam success.

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Mass Percent and Mixture Composition Quiz

17 MCQs

Topics Covered: Composition of Mixtures, Mass Percent, Elemental Analysis, Mixture Purity

Description: This Topic 1.4 quiz tests your understanding of mixture composition. You will practice determining the mass percent of specific elements and assessing the purity of a substance based on elemental analysis.

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1. A sample of C₂H₄ (molar mass 28 g/mol) is contaminated with an unknown impurity. The mass percent of carbon in the impure sample is determined to be 89.5%. Pure C₂H₄ is 85.7% carbon by mass. Which of the following could be the impurity?

CH₄ (75% C)

C₆H₆ (92.3% C)

H₂O (0% C)

CO₂ (27.3% C)

Explanation: Since the experimental mass percent of carbon (89.5%) is higher than the expected mass percent in pure C₂H₄ (85.7%), the impurity must have a carbon mass percent higher than 85.7%. Among the choices, only C₆H₆ (72/78 = 92.3%) has a higher carbon mass percent.

Explanation: Since the experimental mass percent of carbon (89.5%) is higher than the expected mass percent in pure C₂H₄ (85.7%), the impurity must have a carbon mass percent higher than 85.7%. Among the choices, only C₆H₆ (72/78 = 92.3%) has a higher carbon mass percent.

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2. A student is given a mixture containing only solid Cu and solid Zn (brass). The student reacts a 2.00 g sample of the mixture with excess HCl(aq), which reacts completely with the Zn but leaves the Cu unreacted. If 1.30 g of unreacted Cu is recovered, what was the mass percent of Zn in the original sample?

130%

70.0%

65.0%

35.0%

Explanation: The mass of Zn that reacted is the total mass minus the unreacted Cu: 2.00 g - 1.30 g = 0.70 g Zn. The mass percent of Zn is (0.70 g / 2.00 g) x 100 = 35.0%.

Explanation: The mass of Zn that reacted is the total mass minus the unreacted Cu: 2.00 g - 1.30 g = 0.70 g Zn. The mass percent of Zn is (0.70 g / 2.00 g) x 100 = 35.0%.

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3. A mixture consists of two gases, He and Ne, in equal masses. Which of the following statements about this mixture is true?

The mole fraction of He is greater than the mole fraction of Ne.

The mole fraction of Ne is greater than the mole fraction of He.

The mixture contains a higher mass percent of Ne than He.

The mixture contains an equal number of moles of He and Ne.

Explanation: Because He (4.00 g/mol) is much lighter than Ne (20.18 g/mol), equal masses mean there are many more moles of He than moles of Ne. Therefore, the mole fraction of He is greater than that of Ne.

Explanation: Because He (4.00 g/mol) is much lighter than Ne (20.18 g/mol), equal masses mean there are many more moles of He than moles of Ne. Therefore, the mole fraction of He is greater than that of Ne.

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4. A 5.00 g sample of a mixture containing only CaCO₃ (molar mass 100 g/mol) and an inert, non-reactive impurity is heated to decompose the CaCO₃ into CaO and CO₂ gas. If the mass of the solid sample decreases by 1.10 g after heating, what is the mass percent of CaCO₃ in the original mixture?

50.0%

56.0%

44.0%

22.0%

Explanation: The 1.10 g mass lost is CO₂ gas. Moles of CO₂ = 1.10 g / 44.0 g/mol = 0.025 mol CO₂. Because 1 mol CaCO₃ produces 1 mol CO₂, there were 0.025 mol CaCO₃ in the sample. Mass of CaCO₃ = 0.025 mol x 100 g/mol = 2.50 g. Mass percent CaCO₃ = (2.50 g / 5.00 g) x 100 = 50.0%.

Explanation: The 1.10 g mass lost is CO₂ gas. Moles of CO₂ = 1.10 g / 44.0 g/mol = 0.025 mol CO₂. Because 1 mol CaCO₃ produces 1 mol CO₂, there were 0.025 mol CaCO₃ in the sample. Mass of CaCO₃ = 0.025 mol x 100 g/mol = 2.50 g. Mass percent CaCO₃ = (2.50 g / 5.00 g) x 100 = 50.0%.

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5. In a laboratory experiment, a student determines the mass percent of silver in a sample of AgCl by precipitating it from a solution of an unknown mixture. The student forgets to completely dry the AgCl precipitate before determining its mass. How will this error affect the calculated mass percent of silver in the original mixture?

It will be artificially low.

It will remain unchanged because water does not contain silver.

It cannot be determined without knowing the volume of water.

It will be artificially high.

Explanation: By failing to dry the precipitate, the final measured mass of the "AgCl" will include the mass of water. The student will assume this entire mass is AgCl, artificially inflating the calculated moles of AgCl, and thereby calculating a higher mass of silver than is actually present in the original sample.

Explanation: By failing to dry the precipitate, the final measured mass of the "AgCl" will include the mass of water. The student will assume this entire mass is AgCl, artificially inflating the calculated moles of AgCl, and thereby calculating a higher mass of silver than is actually present in the original sample.

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6. A mixture of NO gas and NO₂ gas is placed in a rigid container. The mass percent of nitrogen in pure NO is 46.7%, and the mass percent of nitrogen in pure NO₂ is 30.4%. Elemental analysis of the gas mixture reveals that it is 40.0% nitrogen by mass. Which gas is present in a greater mass in the mixture?

NO, because the mixture's nitrogen percentage is closer to that of NO than NO₂.

They are present in equal masses because 40.0% is roughly halfway between.

It cannot be determined without knowing the volume of the container.

NO₂, because the mixture's nitrogen percentage is closer to that of NO₂ than NO.

Explanation: The mixture's nitrogen mass percent (40.0%) is closer to the mass percent of pure NO (46.7%) than to pure NO₂ (30.4%). Therefore, the mixture must contain a larger mass of NO to skew the average closer to NO's value.

Explanation: The mixture's nitrogen mass percent (40.0%) is closer to the mass percent of pure NO (46.7%) than to pure NO₂ (30.4%). Therefore, the mixture must contain a larger mass of NO to skew the average closer to NO's value.

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7. Two unlabelled white powders are known to be pure BaCl₂ and a mixture of BaCl₂ and NaCl. Elemental analysis of the pure BaCl₂ shows it is 65.9% barium by mass. The unlabelled mixture will have a barium mass percent that is:

unpredictable without knowing the exact mole ratio of the mixture.

exactly 65.9% because barium is the heaviest element in the mixture.

greater than 65.9% because NaCl has a lower molar mass than BaCl₂.

less than 65.9% because the presence of NaCl adds mass to the sample without adding any barium.

Explanation: NaCl contains no barium. Adding it to pure BaCl₂ increases the total mass of the mixture (the denominator) without increasing the mass of the barium (the numerator), strictly decreasing the mass percent of barium overall.

Explanation: NaCl contains no barium. Adding it to pure BaCl₂ increases the total mass of the mixture (the denominator) without increasing the mass of the barium (the numerator), strictly decreasing the mass percent of barium overall.

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8. A student mixes 10.0 g of NaCl (39.3% Na by mass) and 10.0 g of KCl (52.4% K by mass). What is the approximate mass percent of chloride (Cl) in the total 20.0 g mixture?

60.7%

47.6%

100%

54.1%

Explanation: NaCl is 100% - 39.3% = 60.7% Cl. Mass of Cl in NaCl = 0.607 x 10.0 g = 6.07 g. KCl is 100% - 52.4% = 47.6% Cl. Mass of Cl in KCl = 0.476 x 10.0 g = 4.76 g. Total mass of Cl = 6.07 g + 4.76 g = 10.83 g. Mass percent of Cl in mixture = (10.83 g / 20.0 g) x 100 = 54.15%.

Explanation: NaCl is 100% - 39.3% = 60.7% Cl. Mass of Cl in NaCl = 0.607 x 10.0 g = 6.07 g. KCl is 100% - 52.4% = 47.6% Cl. Mass of Cl in KCl = 0.476 x 10.0 g = 4.76 g. Total mass of Cl = 6.07 g + 4.76 g = 10.83 g. Mass percent of Cl in mixture = (10.83 g / 20.0 g) x 100 = 54.15%.

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9. A sample of an unknown alkali metal chloride, MCl, is analyzed. The mass percent of chlorine in the sample is found to be 60.6%. Which of the following is the identity of the alkali metal, M?

Rb

Na

K

Li

Explanation: Let molar mass of M be 'x'. The molar mass of MCl is x + 35.45. Mass percent of Cl = 35.45 / (x + 35.45) = 0.606. Solving for x: 35.45 = 0.606x + 21.48. 13.97 = 0.606x. x = 23.05 g/mol. This corresponds to the molar mass of Sodium (Na).

Explanation: Let molar mass of M be 'x'. The molar mass of MCl is x + 35.45. Mass percent of Cl = 35.45 / (x + 35.45) = 0.606. Solving for x: 35.45 = 0.606x + 21.48. 13.97 = 0.606x. x = 23.05 g/mol. This corresponds to the molar mass of Sodium (Na).

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10. Magnesium reacts with oxygen to form MgO. A student is analyzing a sample of powdered magnesium that has been partially oxidized in the air, creating a mixture of Mg and MgO. The mass percent of oxygen in the mixture is found to be 20.0%. The mass percent of oxygen in pure MgO is approximately 40.0%. What is the mass percent of pure Mg in the mixture?

20.0%

60.0%

80.0%

50.0%

Explanation: If the overall mixture is 20.0% oxygen, the remaining mass must be magnesium. Therefore, the total mixture is 100% - 20.0% = 80.0% magnesium by mass.

Explanation: If the overall mixture is 20.0% oxygen, the remaining mass must be magnesium. Therefore, the total mixture is 100% - 20.0% = 80.0% magnesium by mass.

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11. The particulate-level diagram of a substance shows two different types of diatomic molecules, X₂ and Y₂, uniformly distributed in a container. This representation best describes:

A homogeneous mixture of two elements.

A pure compound of formula XY.

A pure element in a diatomic state.

A heterogeneous mixture of two compounds.

Explanation: Because the molecules are X₂ and Y₂, they are discrete elements. Because both exist in the same container without being chemically bonded to one another (not XY), it is a mixture. Uniform distribution implies a homogeneous mixture (gas mixture).

Explanation: Because the molecules are X₂ and Y₂, they are discrete elements. Because both exist in the same container without being chemically bonded to one another (not XY), it is a mixture. Uniform distribution implies a homogeneous mixture (gas mixture).

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12. A 10.0 g sample of a mixture of sand (SiO₂) and salt (NaCl) is dissolved in water, filtered, and evaporated to dryness. If the dried salt has a mass of 3.5 g, what is the mass percent of sand in the original mixture?

35%

100%

70%

65%

Explanation: The mass of the salt in the mixture is 3.5 g. The mass of the sand must be 10.0 g - 3.5 g = 6.5 g. The mass percent of the sand is (6.5 g / 10.0 g) x 100 = 65%.

Explanation: The mass of the salt in the mixture is 3.5 g. The mass of the sand must be 10.0 g - 3.5 g = 6.5 g. The mass percent of the sand is (6.5 g / 10.0 g) x 100 = 65%.

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13. An impure sample of KClO₃ (molar mass 122.5 g/mol) is decomposed by heating, producing KCl and O₂ gas. A 10.0 g sample of the impure mixture produces 1.92 g of O₂. If the impurity does not decompose, what is the approximate purity (mass percent) of the KClO₃ in the sample?

24.5%

49.0%

98.0%

19.2%

Explanation: 1.92 g O₂ is 1.92 / 32 = 0.060 mol O₂. The reaction is 2 KClO₃ -> 2 KCl + 3 O₂. Moles KClO₃ = 0.060 x (2/3) = 0.040 mol KClO₃. Mass of KClO₃ = 0.040 mol x 122.5 g/mol = 4.90 g. Mass percent = (4.90 g / 10.0 g) x 100 = 49.0%.

Explanation: 1.92 g O₂ is 1.92 / 32 = 0.060 mol O₂. The reaction is 2 KClO₃ -> 2 KCl + 3 O₂. Moles KClO₃ = 0.060 x (2/3) = 0.040 mol KClO₃. Mass of KClO₃ = 0.040 mol x 122.5 g/mol = 4.90 g. Mass percent = (4.90 g / 10.0 g) x 100 = 49.0%.

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14. Two pure compounds, CO and CO₂, are mixed together. The mass percent of carbon in pure CO is 42.9%, and in pure CO₂ is 27.3%. If a mixture of the two gases is found to be 30.0% carbon by mass, what can be concluded about the composition of the mixture?

The mixture contains a higher mass of CO₂ than CO.

The mixture contains equal masses of CO and CO₂.

The mass of CO₂ must be more than double the mass of CO.

The mixture contains a higher mass of CO than CO₂.

Explanation: The experimental mass percent of carbon in the mixture (30.0%) is much closer to the mass percent of carbon in pure CO₂ (27.3%) than to pure CO (42.9%). Therefore, CO₂ must make up the majority of the mixture's mass.

Explanation: The experimental mass percent of carbon in the mixture (30.0%) is much closer to the mass percent of carbon in pure CO₂ (27.3%) than to pure CO (42.9%). Therefore, CO₂ must make up the majority of the mixture's mass.

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15. A pure sample of solid NaCl (molar mass 58.44 g/mol) contains 39.3% sodium by mass. A student analyzes a second, impure sample of NaCl and finds the mass percent of sodium to be 32.1%. Which of the following could be the identity of the impurity in the second sample?

Na₂O (molar mass 62.0 g/mol)

NaOH (molar mass 40.0 g/mol)

LiCl (molar mass 42.4 g/mol)

NaF (molar mass 42.0 g/mol)

Explanation: The impure sample has a lower mass percent of sodium (32.1%) than pure NaCl (39.3%). To lower the mass percent of sodium in the mixture, the impurity must either contain no sodium or have a mass percent of sodium lower than 39.3%. NaF, Na₂O, and NaOH all have higher mass percentages of sodium than NaCl. LiCl contains no sodium, which would decrease the overall mass percent of sodium in the mixture.

Explanation: The impure sample has a lower mass percent of sodium (32.1%) than pure NaCl (39.3%). To lower the mass percent of sodium in the mixture, the impurity must either contain no sodium or have a mass percent of sodium lower than 39.3%. NaF, Na₂O, and NaOH all have higher mass percentages of sodium than NaCl. LiCl contains no sodium, which would decrease the overall mass percent of sodium in the mixture.

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16. A 100 g sample of an alloy contains 80 g of Gold (Au) and 20 g of Silver (Ag). What is the mole fraction of Silver in the alloy? (Molar mass of Au = 197 g/mol, Ag = 108 g/mol)

0.457

0.314

0.686

0.200

Explanation: Moles Au = 80 g / 197 g/mol = 0.406 mol. Moles Ag = 20 g / 108 g/mol = 0.185 mol. Total moles = 0.591 mol. Mole fraction of Ag = 0.185 / 0.591 = 0.313 (approx 0.314). (Note: While standard questions ask for mass %, AP often links mass % to mole ratios in mixtures).

Explanation: Moles Au = 80 g / 197 g/mol = 0.406 mol. Moles Ag = 20 g / 108 g/mol = 0.185 mol. Total moles = 0.591 mol. Mole fraction of Ag = 0.185 / 0.591 = 0.313 (approx 0.314). (Note: While standard questions ask for mass %, AP often links mass % to mole ratios in mixtures).

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17. A 2.50 g sample of an impure iron ore, primarily Fe₂O₃ (molar mass 159.7 g/mol), is reduced completely to solid iron (Fe). The mass of the pure iron recovered is 1.40 g. Assuming the impurities do not contain iron, what is the mass percent of Fe₂O₃ in the original ore sample?

70.0%

100%

80.0%

56.0%

Explanation: The recovered Fe is 1.40 g. Moles of Fe = 1.40 g / 55.85 g/mol = 0.025 mol Fe. Since Fe₂O₃ has 2 moles of Fe per mole of compound, moles of Fe₂O₃ = 0.0125 mol. Mass of Fe₂O₃ = 0.0125 mol x 159.7 g/mol = ~2.00 g. Mass percent Fe₂O₃ = (2.00 g / 2.50 g) x 100 = 80.0%.

Explanation: The recovered Fe is 1.40 g. Moles of Fe = 1.40 g / 55.85 g/mol = 0.025 mol Fe. Since Fe₂O₃ has 2 moles of Fe per mole of compound, moles of Fe₂O₃ = 0.0125 mol. Mass of Fe₂O₃ = 0.0125 mol x 159.7 g/mol = ~2.00 g. Mass percent Fe₂O₃ = (2.00 g / 2.50 g) x 100 = 80.0%.

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Mixtures Are Not Compounds

A pure substance has a fixed composition. Water is always 11.2% hydrogen by mass, and calcium carbonate is always 40.0% calcium, because their formulas fix the ratios.

A mixture has no formula. It is two or more substances physically combined, and its composition depends on how much of each part went in. Seawater, air, alloys and ores are all mixtures, and no two samples need match exactly.

Topic 1.4 is about describing that composition precisely, and about moving between the mass of a mixture, the mass of one component inside it, and the mass of one element inside that component.

Mass percent of a component. Mass percent = (mass of the component ÷ total mass of the mixture) × 100. Every mass percent in this topic is a mass ratio, never a mole ratio, unless a question says otherwise.

Worked Example: Mass Percent in Both Directions

A salt solution is made by dissolving 18.0 g of sodium chloride in water to give 120.0 g of solution in total.

Finding the percent. Mass percent = 18.0 ÷ 120.0 × 100 = 15.0%. Note the denominator: the mass of the whole solution, water included, not the mass of water alone.

Running it in reverse. How much of this solution contains 5.00 g of salt? Mass of solution = 5.00 ÷ 0.150 = 33.3 g. A percent is a conversion factor between component mass and mixture mass, and it works in both directions.

A second reverse. A brass alloy is 65.0% zinc by mass. A 240 g fitting therefore holds 0.650 × 240 = 156 g of zinc, and 84 g of copper. The components’ percentages must sum to 100, so one percent gives the other free.

The Tracer Element Method

Exam questions rarely hand you the mass of a component directly. More often they give the mass of one element and expect you to work back to the component that contains it.

The setup. This works when the element appears in only one component of the mixture. That element is a tracer: every atom of it in the sample came from that one component.

The two steps. First, find what fraction of the component’s mass is the tracer element, using its formula: fraction = (atoms of the element × its molar mass) ÷ (molar mass of the component). Second, divide the measured element mass by that fraction to get the component mass.

Worked example. A 5.00 g rock sample is a mixture of calcium carbonate and sand, SiO₂. Analysis finds 1.20 g of calcium in the sample. Sand contains no calcium, so calcium is the tracer for CaCO₃.

The fraction of Ca in CaCO₃ is 40.08 ÷ 100.09 = 0.4004. The mass of CaCO₃ is therefore 1.20 ÷ 0.4004 = 3.00 g, and the sample is 3.00 ÷ 5.00 × 100 = 59.9% calcium carbonate by mass.

Ores, Purity and Impure Samples

Ores. A metal ore is a mixture of a metal compound and waste rock. Finding the mass of metal in an ore takes the tracer method and one more step: percent of the compound in the ore, then the element fraction inside the compound.

Worked example. A 2.50 g iron ore sample is 68.0% Fe₂O₃ by mass. The mass of Fe₂O₃ is 0.680 × 2.50 = 1.70 g. Iron’s fraction inside Fe₂O₃ is 2(55.85) ÷ 159.70 = 0.6994. The mass of iron is 1.70 × 0.6994 = 1.19 g.

Purity. Percent purity is mass percent by another name: (mass of the pure substance ÷ mass of the impure sample) × 100. A reagent that is 92.0% pure delivers only 0.920 g of the substance per gram weighed out.

Purity matters before any mole calculation. Convert the weighed mass to the pure mass first, then to moles. Moles calculated from the impure mass are always too large.

Drying and mass loss. When a wet sample is dried, the mass lost is the water. A 2.40 g soil sample that dries to 2.04 g lost 0.36 g of water, so the original sample was 0.36 ÷ 2.40 × 100 = 15.0% water by mass. The denominator is the original, wet mass.

Traps That Cost Marks in Topic 1.4

Reporting the compound’s mass as the element’s mass. “The sample contains 1.70 g of Fe₂O₃” is not “the sample contains 1.70 g of iron”. The element is only a fraction of its compound, 0.6994 here, and stopping one step early is the most common error in this topic.

Dividing by the wrong total. Mass percent uses the mass of the whole mixture, solution or sample, including every component and the solvent. Using just the water, or just the other component, as the denominator breaks the definition.

Choosing a tracer that is not unique. The tracer method fails if the element appears in two components. Oxygen, for example, can almost never be a tracer in a mixture of oxides, because every component contains it.

Forgetting purity before moles. A question that gives an impure mass and asks for moles, particles or a reacting mass expects the purity correction first. The impurity is inert filler; it contributes mass and nothing else.

Confusing mass percent with mole percent. Composition data in this topic is by mass. Two components at 50% by mass are not present in equal moles unless their molar masses happen to match.

Using the dry mass as the denominator for water content. Percent water in a wet sample is water mass divided by the original wet mass. Dividing by the dried mass answers a different question.

Key Relationships

• Mass percent of a component = (mass of component ÷ total mass of mixture) × 100; all components’ percents sum to 100
• Component mass = mass percent (as a decimal) × total mixture mass
• Tracer method: mass of component = mass of tracer element ÷ (mass fraction of the element in the component), valid only when the element occurs in just one component
• Mass fraction of an element in a compound = (atoms × atomic mass) ÷ molar mass of the compound
• Pure mass = percent purity (as a decimal) × impure sample mass; apply purity before converting to moles
• Percent lost on drying = (mass lost ÷ original mass) × 100